Question

A mango in a tree is located at a horizontal and vertical distance of 30 m and 40 m respectively from the point of projection of a stone. Find the minimum speed of the stone so as to hit the mango,
A. u > 30m/s
B. u < 30m/s
C. u = 50m/s
D. u = 40m/s

Answer 1

Hint: The motion of a projectile is a two dimensional motion.
It is separated into two components u cosƟ and u sinƟ.
At a maximum height, initial velocity in downward direction is u sinƟ.

Complete step by step answer:
 A stone is being projected to hit a mango in a tree which is located at a horizontal distance of h = 30 m and vertical distance of y = 40 m and let the stone be thrown with a velocity of u at an angle of Ɵ with the horizontal. The velocity u is split into its two components u cosƟ and u sinƟ. The horizontal motion takes place with constant velocity u cosƟ and the distance h covered is in time t.
We can assume that the mango is at its maximum height from the ground and maximum height is 40m.
From kinematic equation, $v = u\sin \theta - gt[v = u + at]$
$0 = u\sin \theta - gt$ [Since the velocity at highest point is zero]
$u\sin \theta = gt$
$t = \dfrac{{u\sin \theta }}{g}$
$y = \left( {u\sin \theta } \right)t - \dfrac{1}{2}g{t^2}\left[ {y = ut + \dfrac{1}{2}a{t^2}} \right] \\ $
$y = ut + \dfrac{1}{2}g{t^2} $
$\implies y = \left( {u\sin \theta } \right)\left( {\dfrac{{u\sin \theta }}{g}} \right) - \dfrac{1}{2}g{\left( {\dfrac{{u\sin \theta }}{g}} \right)^2} \\ $
$\implies y = \dfrac{{{u^2}{{\sin }^2}\theta }}{g} - \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\$
$\implies y = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}} \\ $
$u = \sqrt {\dfrac{{2yg}}{{{{\sin }^2}\theta }}} $
For minimum speed, the value of sinƟ is maximum and therefore the value of Ɵ is 90⁰. Hence, sin90⁰=1.
$ u = \sqrt {\dfrac{{2 \times 40 \times 10}}{{{1^2}}}} $
$ u = \sqrt {800} = 28.28m{s^{ - 1}}[approx.value] $
Thus, minimum velocity required to hit the mango is less than $30m{s^{ - 1}}$ .

So, the correct answer is “Option b”.

Note:
The initial velocity at the highest point is due to the y component of velocity i.e., v sinƟ.
When the stone reaches the highest point, the final velocity becomes zero as kinetic energy is zero.